Answer:
(-2, -8)
x = -2
y = -8
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define systems</u>
13x - 6y = 22
x = y + 6
<u>Step 2: Solve for </u><em><u>y</u></em>
<em>Substitution</em>
- Substitute in <em>x</em>: 13(y + 6) - 6y = 22
- Distribute 13: 13y + 78 - 6y = 22
- Combine like terms: 7y + 78 = 22
- Isolate <em>y</em> term: 7y = -56
- Isolate <em>y</em>: y = -8
<u>Step 3: Solve for </u><em><u>x</u></em>
- Define original equation: x = y + 6
- Substitute in <em>y</em>: x = -8 + 6
- Add: x = -2
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Answer: 
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Given: 
Find: 
Solution: In order to get the equation, we need to first use the point-slope form to help us sort out our information and then we solve for y.
<u>Plug in the values</u>
<u>Simplify and solve for y</u>
Therefore, the final answer would be that the equation that has a slope of -1/4 and passes through (4, -5) is y = -1/4x - 4.
Answer:
1
Step-by-step explanation:
You can only cut it in half (vertically)
Answer:
1067
Step-by-step explanation:
Given that:
Error (E) = 0.03
Confidence interval = 95% = 0.95
Sample size (n) = ((Zα/2) / Error)^2 * p(1 - p)
Since no assumption is given ; p = 0.5
α = 0.95
Z(1-α/2) = 1.96
(1.96 / 0.03)^2 * 0.5(1 - 0.5)
(65.333333)^2 * 0.5(0.5)
4268.4444 * 0.25
= 1067.1111
n = 1067
let's notice something, the parabola is a vertical one, so the squared variable is the x, and is opening downwards, meaning the x² will have a negative coefficient.
the distance from the vertex to the directrix/focus is the amount of "p" units, let's see in the graph, the distance from the vertex to the directrix is 2, and since the parabola is opening downwards, "p" is a negative 2, p = -2. The vertex is of course at (0, 2).
![\bf \textit{parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=2\\ p=-2 \end{cases}\implies 4(-2)(y-2)=(x-0)^2\implies -8(y-2)=x^2 \\\\\\ y-2=\cfrac{x^2}{-8}\implies \blacktriangleright y=-\cfrac{1}{8}x^2+2 \blacktriangleleft](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%0A%5C%5C%5C%5C%0A4p%28y-%20k%29%3D%28x-%20h%29%5E2%0A%5Cqquad%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%20%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ah%3D0%5C%5C%0Ak%3D2%5C%5C%0Ap%3D-2%0A%5Cend%7Bcases%7D%5Cimplies%204%28-2%29%28y-2%29%3D%28x-0%29%5E2%5Cimplies%20-8%28y-2%29%3Dx%5E2%0A%5C%5C%5C%5C%5C%5C%0Ay-2%3D%5Ccfrac%7Bx%5E2%7D%7B-8%7D%5Cimplies%20%5Cblacktriangleright%20y%3D-%5Ccfrac%7B1%7D%7B8%7Dx%5E2%2B2%20%5Cblacktriangleleft%20)